---
title: "20 kVA 250/2500 V tests, equivalent circuit"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-20kva-250-2500-parameters"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-20kva-250-2500-parameters
site: BCT Engineering
---

# 20 kVA 250/2500 V tests, equivalent circuit

Electrical Machines · Chapter 2: Transformer

- Asked in: 2072 Asoj
- Video: [EE213 - 09b - Determining equivalent circuit parameters for a transformer - numerical example](https://www.youtube.com/watch?v=b_6NbUb7U1o) — MAFarooqi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-20kva-250-2500-parameters
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 20 kVA, 250 V/2500 V, 50 Hz single phase transformer gave the following test results:
No-load test (on L.V. side): 250 V, 1.4 A, 105 watts
Short circuit test (on H.V. side): 120 V, 8 A, 320 watts
Calculate the equivalent circuit parameters referred to primary side and draw the equivalent circuit.

## Short answer

Equivalent circuit referred to primary: shunt branch R0=595.2 Ω, X0=187.2 Ω; series branch R01=0.05 Ω, X01=0.1414 Ω.

## Answer

**Given:** 20 kVA, 250 V/2500 V. Primary = LV (250 V), secondary = HV (2500 V), $K = 10$, $K^2 = 100$.

#### Open-circuit test (on the LV side)

The no-load power is almost entirely core loss, so the shunt branch is found from it.

$$
\begin{aligned}
\cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{105}{250\times1.4} = 0.3 \\
I_w &= I_0\cos\phi_0 = 0.42\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.336\ \text{A} \\
R_0 &= \frac{V_0}{I_w} = 595.2\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 187.2\ \Omega\quad(\text{on the LV side})
\end{aligned}
$$

#### Short-circuit test (on the HV side)

At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss.

$$
\begin{aligned}
Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\
R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\
X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 14.14\ \Omega\quad(\text{on the HV side})
\end{aligned}
$$

#### Referring to the LV side

The turns ratio is $K = 2500/250 = 10$, so $K^2 = 100$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$
\begin{aligned}
R_{eq} &= 5 \div 100 = 0.05\ \Omega \\ X_{eq} &= 14.14 \div 100 = 0.1414\ \Omega
\end{aligned}
$$

#### Equivalent circuit referred to the LV side

```text
   Req=0.05 ohm   Xeq=0.141 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=595.2] [X0=187.2]   Load
    |      (in parallel)
 o--+---------------------------o
```

$R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$ ($Z_{eq} = 0.15\ \Omega$), shunt branch $R_0 = 595.2\ \Omega$ in parallel with $X_0 = 187.2\ \Omega$.

**Answer (referred to primary):** $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which side is the OC test done on, and what does it give?
   A. LV side; shunt branch R0 and X0
   B. HV side; series branch
   C. LV side; series branch
   D. HV side; copper loss
   Answer: A. No-load power is core loss, which determines the shunt branch. (Key point: OC test side)

2. How is the magnetising current found (I0 = 1.4 A, Iw = 0.42 A)?
   A. I0 - Iw = 0.98 A
   B. sqrt(I0^2 - Iw^2) = 1.336 A
   C. I0 + Iw = 1.82 A
   D. I0 x Iw = 0.588 A
   Answer: B. Iw and Imu are at right angles. (Key point: Magnetising current)

3. Why are SC values divided by 100 to refer to the primary?
   A. K = 10 and HV to LV needs division by K^2
   B. Multiply by K^2 is needed
   C. K = 100
   D. Because power is doubled
   Answer: A. Impedance moves from HV to LV by dividing by K^2. (Key point: Referring SC)

## Close variants set in other exams

- [25 kVA 250/2500 V tests, equivalent circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-25kva-250-2500-parameters) — Electrical Machines, chapter 2, asked 2×

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